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Question Number 175548 by Rasheed.Sindhi last updated on 02/Sep/22

determinant ((( determinant (((2+424+44244+4442444+∙∙∙n terms=?_ ^ _() ^() )))_ ^ ^(∣•∣_(−) ^(−) ) )))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\begin{array}{|c|}{\overset{\underset{−} {\overline {\mid\bullet\mid}}} {\:\begin{array}{|c|}{\underset{} {\overset{} {\mathrm{2}+\mathrm{424}+\mathrm{44244}+\mathrm{4442444}+\centerdot\centerdot\centerdot{n}\:{terms}=?_{} ^{} }}}\\\hline\end{array}_{} ^{} }}\\\hline\end{array} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by infinityaction last updated on 02/Sep/22

(2+2)+(424+20)+(44244+200)+...n term −(2+20+200+...+n term) S = { 4+444+44444+...n term −2(1+10+100+..n term)} 4+444+44444+.....n term_(Ψ) − (2/9)(10^n −1) Ψ = 4(1+111+11111+...n term) 9Ψ = 4(9+999 +... n term) Ψ = (4/9)[{10−1}+{10^3 −1}+...n term] Ψ =(4/9)[((10)/(99))×(10^(2n) −1)−n] S = Ψ−(2/9)(10^n −1) S = (4/9)[((10)/(99))×(10^(2n) −1)−n]−(2/9)(10^n −1)

$$\left(\mathrm{2}+\mathrm{2}\right)+\left(\mathrm{424}+\mathrm{20}\right)+\left(\mathrm{44244}+\mathrm{200}\right)+...{n}\:{term} \\ $$$$−\left(\mathrm{2}+\mathrm{20}+\mathrm{200}+...+{n}\:{term}\right) \\ $$$$\:\:\:{S}\:\:=\:\:\left\{\:\mathrm{4}+\mathrm{444}+\mathrm{44444}+...{n}\:{term}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\left(\mathrm{1}+\mathrm{10}+\mathrm{100}+..{n}\:{term}\right)\right\} \\ $$$$\:\:\:\:\underset{\Psi} {\underbrace{\mathrm{4}+\mathrm{444}+\mathrm{44444}+.....{n}\:{term}}}\:−\:\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\:\:\Psi\:=\:\mathrm{4}\left(\mathrm{1}+\mathrm{111}+\mathrm{11111}+...{n}\:{term}\right) \\ $$$$\:\:\:\:\mathrm{9}\Psi\:=\:\mathrm{4}\left(\mathrm{9}+\mathrm{999}\:+...\:{n}\:{term}\right) \\ $$$$\:\:\Psi\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\left\{\mathrm{10}−\mathrm{1}\right\}+\left\{\mathrm{10}^{\mathrm{3}} −\mathrm{1}\right\}+...{n}\:{term}\right]\: \\ $$$$\:\:\:\:\Psi\:=\frac{\mathrm{4}}{\mathrm{9}}\left[\frac{\mathrm{10}}{\mathrm{99}}×\left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right)−{n}\right] \\ $$$$\:\:\:\:\:{S}\:\:\:=\:\:\Psi−\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$$$\:\:\:\:{S}\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\frac{\mathrm{10}}{\mathrm{99}}×\left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right)−{n}\right]−\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{10}^{{n}} −\mathrm{1}\right) \\ $$

Commented by Rasheed.Sindhi last updated on 03/Sep/22

•∩i⊂∈! •Thank you sir!

$$\bullet\cap\boldsymbol{\mathrm{i}}\subset\in!\:\:\:\:\:\:\:\: \\ $$$$\bullet\mathcal{T}{hank}\:{you}\:{sir}! \\ $$

Commented by infinityaction last updated on 03/Sep/22

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Commented by peter frank last updated on 05/Sep/22

thanks

$$\mathrm{thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 04/Sep/22

AnOther way... 2+424+44244+4442444+∙∙∙n terms =(404+44044+4440444+..._(n−1 terms) )+(2+20+200+..._(n terms) ) =(4/9)(909+99099+9990999..._(n−1 terms) )+((2(10^n −1))/((10−1))) =(4/9)(1000−91+100000−901+10000000−9001..._(n−1 terms) )+((2(10^n −1))/9) =(4/9)(10^3 −91+10^5 −901+10^7 −9001..._(n−1 terms) )+((2(10^n −1))/((10−1))) =(4/9){(10^3 +10^5 +10^7 +..._(n−1 terms) )−(90+900+9000+..._(n−1 terms) )−(n−1)}+((2(10^n −1))/((10−1))) =(4/9){((10^3 (100^(n−1) −1))/(100−1))−((90(10^(n−1) −1))/(10−1))−n+1}+((2(10^n −1))/((10−1))) =(4/9){((10^3 (100^(n−1) −1))/(99))−((90(10^(n−1) −1))/9)−n+1}+((2(10^n −1))/9) =(4/9){((10^3 (100^(n−1) −1)−990(10^(n−1) −1)−99n+99)/(99))}+((2(10^n −1))/9) =(4/9){((10^(2n+1) −990∙10^(n−1) −99n+89)/(99))}+((2(10^n −1))/9) =(4/9){((10∙10^(2n) −10−99∙10^n −99n+99)/(99))}+((2(10^n −1))/9) =(4/9){((10)/(99))(10^(2n) −1)−n}−(2/9)((10^n −1))

$$\boldsymbol{\mathrm{AnOther}}\:\boldsymbol{\mathrm{way}}... \\ $$$$\mathrm{2}+\mathrm{424}+\mathrm{44244}+\mathrm{4442444}+\centerdot\centerdot\centerdot{n}\:{terms} \\ $$$$=\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{404}+\mathrm{44044}+\mathrm{4440444}+...}}\:\right)+\left(\underset{{n}\:{terms}} {\underbrace{\mathrm{2}+\mathrm{20}+\mathrm{200}+...}}\right) \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{909}+\mathrm{99099}+\mathrm{9990999}...}}\right)+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\left(\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{1000}−\mathrm{91}+\mathrm{100000}−\mathrm{901}+\mathrm{10000000}−\mathrm{9001}...}}\right)+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{10}^{\mathrm{3}} −\mathrm{91}+\mathrm{10}^{\mathrm{5}} −\mathrm{901}+\mathrm{10}^{\mathrm{7}} −\mathrm{9001}...}}\right)+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\left(\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{10}^{\mathrm{3}} +\mathrm{10}^{\mathrm{5}} +\mathrm{10}^{\mathrm{7}} +...}}\:\right)−\left(\underset{{n}−\mathrm{1}\:{terms}} {\underbrace{\mathrm{90}+\mathrm{900}+\mathrm{9000}+...}}\:\right)−\left({n}−\mathrm{1}\right)\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\left(\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}^{\mathrm{3}} \left(\mathrm{100}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{100}−\mathrm{1}}−\frac{\mathrm{90}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{10}−\mathrm{1}}−{n}+\mathrm{1}\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\left(\mathrm{10}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}^{\mathrm{3}} \left(\mathrm{100}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{99}}−\frac{\mathrm{90}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{1}\right)}{\mathrm{9}}−{n}+\mathrm{1}\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}^{\mathrm{3}} \left(\mathrm{100}^{{n}−\mathrm{1}} −\mathrm{1}\right)−\mathrm{990}\left(\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{1}\right)−\mathrm{99}{n}+\mathrm{99}}{\mathrm{99}}\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}^{\mathrm{2}{n}+\mathrm{1}} −\mathrm{990}\centerdot\mathrm{10}^{{n}−\mathrm{1}} −\mathrm{99}{n}+\mathrm{89}}{\mathrm{99}}\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}\centerdot\mathrm{10}^{\mathrm{2}{n}} −\mathrm{10}−\mathrm{99}\centerdot\mathrm{10}^{{n}} −\mathrm{99}{n}+\mathrm{99}}{\mathrm{99}}\right\}+\frac{\mathrm{2}\left(\mathrm{10}^{{n}} −\mathrm{1}\right)}{\mathrm{9}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{9}}\left\{\frac{\mathrm{10}}{\mathrm{99}}\left(\mathrm{10}^{\mathrm{2}{n}} −\mathrm{1}\right)−{n}\right\}−\frac{\mathrm{2}}{\mathrm{9}}\left(\left(\mathrm{10}^{{n}} −\mathrm{1}\right)\right) \\ $$

Commented by Tawa11 last updated on 15/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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